a ratio that comes from a fraction is the numerator: denominator

7/8 = 7:8

1m 40cm = 140cm. 1m = 100cm. So the ratio is 140cm:100cm. This can be put as a fraction 140/100 and then reduced to 14/10 and further to 7/5. This, in turn, can be rewritten as a ratio as 7:5.

One remote is defective for every 199 non-defective remotes.

Begin by making your life easier: presume that there are  papers on the desk. Immediately, we know that there are  paper clips. Now, if there are  papers, you know that there also must be  greeting cards. Technically you figure this out by using the ratio:

By cross-multiplying you get:

Solving for , you clearly get .

(Many students will likely see this fact without doing the algebra, however. The numbers are rather simple.)

Now, this means that our desk has on it:

 papers

 paper clips

 greeting cards

Therefore, you have  total items.  Based on this, your ratio of paper clips to total items is:

, which is the same as .

To begin, you need to calculate how many students are taking Old Norse. This is:

Now, the ratio of students taking Old Norse to students taking Greek is the same thing as the fraction of students taking Old Norse to students taking Greek, or:

Next, just reduce this fraction to its lowest terms by dividing the numerator and denominator by their common factor of :

This is the same as .

To begin, you need to do a simple addition to find the total number of flowers in the garden:

Now, the ratio of petunias to the total number of flowers in the garden can be represented by a simple division of the number of petunias by . This is:

Next, reduce the fraction by dividing out the common  from the numerator and the denominator:

This is the same as .

We create a conversion ratio that causes yards to cancel out, leaving only cents in the numerator and quilts in the denominator. This ratio is ((c cent )/(10 yard))((1 yard)/(q quilt))=(c )/(10q ) cent⁄quilt . Since the ratio has cents in the numerator and quilts in the denominator, it represents the price in cents per quilt.

1/3 of sales from cupcakes = $20, ¼ of sales from cream pies = $15 and the rest are from brownies = $60-$20-$15 = $25. Since each brownie costs 25 cents, Susan will have sold 100 of them.

Consider

(Bill + 7) = 2 x (Amy + 7)

(Amy – 2) = 1.5 x (Molly – 2)

Solve for Molly using the two equations by finding Amy’s age in terms of Molly’s age.

Amy = 2 + 1.5 Molly – 3 = 1.5 x Molly – 1

Substitute this into the first equation:

(Bill + 7) = 2 x (Amy + 7) = 2 x (1.5 x Molly – 1 + 7) = 2 x (1.5 x Molly + 6) = 3 x Molly + 12

Solve for Molly:

Bill + 7 – 12 = 3 x Molly

Molly = (Bill – 5) ¸ 3

Substitute Bill = 29

Molly = (Bill – 5) ¸ 3 = 8

Let original white chocolate pieces = W and original milk chocolate pieces = M. So the total number of pieces in the original bag is M + W.

From the first sentence: (M + W) x 0.2 = W or

0.2 M = 0.8 W or [M = 4W]

Once Christina has eaten 10 milk chocolate pieces, there are W pieces of white chocolate, (M – 10) pieces of milk chocolate and (M + W – 10) pieces total. According to the second sentence:

W ¸ (M – 10) = 3 ¸ 2

Or 2W = 3M - 30

Insert the equation in brackets: 2W = 3[4W] + 30 = 12W – 30

10W = 30 or W = 3 and M = 12

We want “How many pieces of chocolate are left in the bag” or (M – W – 10).

So (M +W – 10) = 3 + 12 – 10 = 5